Question: Let $S$ be the cylinder with height $5$ and radius $2$ whose axis is parallel to the $z$ -axis and whose lower base is centered at the origin. What is the triple integral of the scalar field $f(x, y, z) = 1$ over $S$ in cylindrical coordinates? Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^5 \int_0^2 \int_0^{2\pi} r \, d\theta \, dr \, dz$ (Choice B) B $ \int_{-5}^5 \int_0^2 \int_0^{2\pi} r \, d\theta \, dr \, dz$ (Choice C) C $ \int_{-5}^5 \int_0^2 \int_0^{2\pi} r^2 \, d\theta \, dr \, dz$ (Choice D) D $ \int_0^5 \int_0^2 \int_0^{2\pi} r^2 \, d\theta \, dr \, dz$
Explanation: The only bound is $0 < \theta < 2\pi$. Here is the change of variables for cylindrical coordinates. $\begin{aligned} x &= r \cos(\theta) \\ \\ y &= r \sin(\theta) \\ \\ z &= z \end{aligned}$ We want to represent the cylinder $S$ with bounds in cylindrical coordinates. Here, the region $S$ ranges from a height of $z = 0$ to $z = 5$ and radius of $r = 0$ to $r = 2$. Theta goes from $0$ to $2\pi$ so that we integrate over the whole cylinder and not just a wedge. $ \int_0^5 \int_0^2 \int_0^{2\pi} \cdots \, d\theta \, dr \, dz$ The scalar field over which we want to integrate is just $1$, so the integral looks the same: $ \int_0^5 \int_0^2 \int_0^{2\pi} \cdots \, d\theta \, dr \, dz$ The final step is finding the Jacobian of cylindrical coordinates, which we'll need to multiply in to get the final integral. $J(r, \theta, z) = r$ [Derivation] The integral in cylindrical coordinates: $ \int_0^5 \int_0^2 \int_0^{2\pi} r \, d\theta \, dr \, dz$